Question 65
Let $a,b,c\gt0$. Prove
$$\frac{a+b+3c}{3a+3b+2c}+\frac{a+3b+c}{3a+2b+3c}+\frac{3a+b+c}{2a+3b+3c}\ge\frac{15}{8}.$$
$$\frac{a+b+3c}{3a+3b+2c}+\frac{a+3b+c}{3a+2b+3c}+\frac{3a+b+c}{2a+3b+3c}\ge\frac{15}{8}.$$
MathNet 2013 Question 65
Math Olympiad Vault · MathNet 2013